First slide

Theory of equations

Question

The number of real solutions of the equation $|x^{2} + 4 x + 3| + 2 x + 5 = 0$ are

Moderate

A

1
B

2
C

3
D

4

Solution

We have, $|x^{2} + 4 x + 3| + 2 x + 5 = 0$
Here, two cases arise.
Case l When x² +4x+3>0
$\begin{array}{r} x^{2} + 4 x + 3 + 2 x + 5 = 0 \\ x^{2} + 6 x + 8 = 0 \\ (x + 2) (x + 4) = 0 \\ x = - 2, - 4 \end{array}$
x = -2 is not satisfying the condition x² + 4x + 3 > 0.
So, x= - 4 is the only solution of the given equation.
Case ll When $x^{2} + 4 x + 3 < 0$
$\begin{array}{lrl} \Rightarrow - (x^{2} + 4 x + 3) + 2 x + 5 = 0 \\ \Rightarrow - x^{2} - 2 x + 2 = 0 \\ \Rightarrow x^{2} + 2 x - 2 = 0 \\ \Rightarrow (x + 1 + \sqrt{3}) (x + 1 - \sqrt{3}) = 0 \\ \Rightarrow x = - 1 + \sqrt{3}, - 1 - \sqrt{3} \end{array}$
Hence, $x = - (1 + \sqrt{3})$ satisfy the given condition. Since, $x^{2} + 4 x + 3 < 0$ while $x = - 1 + \sqrt{3}$ is not satisfying the condition. Thus, number of real solutions are two.

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